IBDP Maths AHL 2.12 Polynomial functions and their graphs AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) Finding the value of \( a \):
By the Remainder Theorem, \( f(a) = -10 \):
\( 4a^3 + 2a^2 – 7a = -10 \)
Rearrange: \( 4a^3 + 2a^2 – 7a + 10 = 0 \)
Factor: \( (a + 2)(4a^2 – 6a + 5) = 0 \)
The quadratic has no real roots (discriminant \( D = (-6)^2 – 4 \times 4 \times 5 = -44 < 0 \))
Thus, the only real solution is \( \boxed{a = -2} \)
(b) Unique real solution for \( f(x) = 0 \):
Substitute \( a = -2 \) into \( f(x) \):
\( f(x) = 4x^3 – 4x + 14 \)
Method 1: Graphical Approach
The graph of \( y = 4x^3 – 4x + 14 \) crosses the x-axis exactly once, indicating a unique real root.
(The function has no local maxima/minima since \( f'(x) = 12x^2 – 4 > 0 \) for all real \( x \))
Method 2: Algebraic Approach
The discriminant for cubic \( x^3 – x + 3.5 = 0 \) is:
\( \Delta = -4(-1)^3 – 27(3.5)^2 = -327.75 < 0 \)
Since \( \Delta < 0 \), there is exactly one real root.
Conclusion: There exists exactly one real solution to \( f(x) = 0 \).
▶️ Answer/Explanation
(a) Showing x=1 is not a root:
Substitute \( x = 1 \) into the quadratic factor:
\( 1^2 + (2-k)(1) + k^2 = 0 \) ⇒ \( k^2 – k + 3 = 0 \)
Discriminant: \( \Delta = (-1)^2 – 4(1)(3) = -11 < 0 \)
Since \( \Delta < 0 \), there are no real \( k \) satisfying the equation.
Conclusion: \( x = 1 \) cannot be a root of the quadratic factor.
(b) Finding values of \( k \):
The quadratic discriminant is: \( \Delta = (2-k)^2 – 4(1)(k^2) = 4 – 4k + k^2 – 4k^2 = -3k^2 -4k +4 \)
(i) Exactly one real root:
Quadratic has no real roots (\( \Delta < 0 \)):
Solve \( -3k^2 -4k +4 < 0 \) ⇒ \( 3k^2 +4k -4 > 0 \)
Roots: \( k = \frac{-4 \pm \sqrt{16+48}}{6} = \frac{-4 \pm 8}{6} \) ⇒ \( k = -2 \) or \( k = \frac{2}{3} \)
Solution: \( \boxed{k < -2} \) or \( \boxed{k > \frac{2}{3}} \)
(ii) Exactly two different real roots:
Quadratic has one real root (\( \Delta = 0 \)):
Solution: \( \boxed{k = -2} \) or \( \boxed{k = \frac{2}{3}} \)
(iii) Three different real roots:
Quadratic has two distinct real roots (\( \Delta > 0 \)):
Solution: \( \boxed{-2 < k < \frac{2}{3}} \)
(c) Roots for case (b)(ii):
When \( k = -2 \):
Quadratic becomes \( x^2 + 4x + 4 = (x+2)^2 \)
Roots: \( \boxed{1} \) (simple root) and \( \boxed{-2} \) (double root)
When \( k = \frac{2}{3} \):
Quadratic becomes \( x^2 + \frac{4}{3}x + \frac{4}{9} = (x+\frac{2}{3})^2 \)
Roots: \( \boxed{1} \) (simple root) and \( \boxed{-\frac{2}{3}} \) (double root)